In the figure above, lines L and P are parallel. The segment AD is the same length as the segment DC; AD is parallel to BC. If the length of AD is equal to 4, and angle ADC is equal to 60º, what is the area of ABCD?

A. 4√2B. 4√3C. 8√2D. 8√3E. 16√2

(D) AB is parallel to DC and AD is parallel to BC, so ABCD is parallelogram. We know how to find the area of a parallelogram we multiply the height of parallelogram by its base length.

We are told that the angle ADC is a 60º angle. If we draw an altitude from A straight down and perpendicular to line P, the length of that altitude will be equal to the height of the rhombus. Moreover, it forms a 60-30-90 triangle, so we can easily find its length. Since the hypotenuse of that triangle is equal to 4, the second longest side must be equal to 2√3 (since the proportions for the triangle run x, x√3, and 2x).

To find the area of a rhombus, multiply the height of the rhombus by its base length: 4 × 2√3 = 8√3, or answer choice (A).-------------The sum of the interior angles should be (n – 2) × 180. So the sum of the interior angles for ABCD = 360⁰. How is it determined that the traingle formed from A is a 30-60-90 triangle?

How did you deduce that the parallelogram formed its special case (i.e., a rhombus)? Given the information from the question and the fact that the figure is not drawn to scale, I find it hard to deduce that the figure is a rhombus.

Question statement tells us:"In the figure above, lines L and P are parallel. ... AD is parallel to BC."So we know that ABCD is a parallelogram. As any parallelogram its opposite sides are equal. So AB = DC and AD = BC.

Besides, "the segment AD is the same length as the segment DC". So now all 4 sides are equal. Therefore ABCD is rhombus.

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